Login

Welcome, Guest. Please login or register.

March 29, 2024, 07:40:43 am

Author Topic: Integration between two curves  (Read 2230 times)

0 Members and 1 Guest are viewing this topic.

shaun.mcvicar

  • Fresh Poster
  • *
  • Posts: 2
  • Respect: 0
Integration between two curves
« on: February 29, 2020, 03:45:30 pm »
0
Hi. Can anyone help with this question?

RuiAce

  • ATAR Notes Lecturer
  • Moderator
  • Great Wonder of ATAR Notes
  • *****
  • Posts: 8814
  • "All models are wrong, but some are useful."
  • Respect: +2575
Re: Integration between two curves
« Reply #1 on: February 29, 2020, 04:10:36 pm »
+1
Which part are you stuck on and what have you tried considering so far?

shaun.mcvicar

  • Fresh Poster
  • *
  • Posts: 2
  • Respect: 0
Re: Integration between two curves
« Reply #2 on: February 29, 2020, 04:17:29 pm »
0
Which part are you stuck on and what have you tried considering so far?
I can't find the upper or lower boundaries for P to even begin the question

RuiAce

  • ATAR Notes Lecturer
  • Moderator
  • Great Wonder of ATAR Notes
  • *****
  • Posts: 8814
  • "All models are wrong, but some are useful."
  • Respect: +2575
Re: Integration between two curves
« Reply #3 on: February 29, 2020, 08:53:52 pm »
+1
I can't find the upper or lower boundaries for P to even begin the question
For region \(P\), you can observe that the "endpoints" of the region are both cut off by the line \(y=e\). For \(y=e^x\), the endpoint at \((1,e)\) is literally plotted for you.

For the left "endpoint", you can simply find the point of intersection between the line \(y=e\) and the curve \(y=e^{-x}\).
\begin{align*}
e &= e^{-x}\\
1 &= -x\\
x &= -1
\end{align*}
So the left "endpoint" is at \((-1, e)\).

However, incidentally enough the easier of the two is the area of region \(Q\) here. For region \(P\), you should identify a region split at \(x=0\), i.e. along the \(y\)-axis.

You should double check the diagram, and observe that:
- For \(x\in [-1,0]\), the upper curve is the line \(y=e\), but the lower curve is \(y=e^{-x}\).
- For \(x\in [0,1]\), the upper curve is the line \(y=e\), but the lower curve is \(y=e^x\).

This results in two separate integrals that you have to evaluate for this problem. You should use the above information to consider what they are.